The answer to this question has essentially been given by @DheerajKumar, but let's explain it in a different way.
You are studying circular motion, that is the motion of a particle along a circular path. In such motion, linear speed $v$ gives you the length of arc covered by the particle per unit of time. With angular speed $\omega$ you want to give a measure of the angle covered by the particle per unit of time.
But, how to define angle? There is an easy way to do it if you notice that the length of arc of any circular path is proportional to the radius $r$ of the circle. Thus you can define the angle covered by the particle as the length of arc covered by it divided by the radius $r$ of the circle. As stated by @DheerajKumar, this is the definition of radian, that is, if you are defining angle in that way, you are measuring it in a unit called radian. With that definition of angle you obtain
$$
\omega = \frac{v}{r},
$$
which is equivalent to
$$
v = \omega \cdot r.
$$
Notice that these formulas are only valid if you use radians as the unit of angle.
To understand the units, notice that $\rm radians$ are $\;\dfrac{meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)}}{meters\ {\rm \scriptstyle (coming\ from\ radius)}}$.
The units of $a_c = \dfrac{v^2}{r}$ are
$$
\frac{
meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)} \cdot meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)}
}
{
meters\ {\rm \scriptstyle (coming\ from\ radius)} \cdot sec \cdot sec
} = {\rm m/s^2}.
$$
You can see is the same thing if you use $a_c = \omega^2 r$ which has the units
$$
\frac
{rad\cdot rad\cdot meters\ {\rm \scriptstyle (coming\ from\ radius)} }
{sec\cdot sec}
=
\frac{
meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)} \cdot meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)}
\cdot meters\ {\rm \scriptstyle (coming\ from\ radius)}
}
{
meters\ {\rm \scriptstyle (coming\ from\ radius)}
\cdot meters\ {\rm \scriptstyle (coming\ from\ radius)}
\cdot sec \cdot sec
} = {\rm m/s^2}.
$$
